# Vectorizing in Python [closed]

I have the following code:

for j in range(rows):
for i in range(cols):
k = j * rows + i

k1 = k + 1 if i < cols - 1 else k - 1
k2 = k - 1 if i > 0 else k + 1
k3 = k + cols if j < rows - 1 else k - cols
k4 = k - cols if j > 0 else k + cols

w1 = U[k]
w2 = U[k-1] if i > 0 else U[k]
w3 = V[k]
w4 = V[k-cols] if j > 0 else V[k]

zarray[k] = (w1 + w2 + w3 + w4) * parray[k] - (w1 * parray[k1] + w2 * parray[k2] + w3 * parray[k3] + w4 * parray[k4])


I want to know if there is a way to vectorize this loops, because I think that exists a kind of "convolution" for zarray.

U and V are arrays representing 2D matrices with 512x512 elements, and parray is also a 2D representation with 512x512 elements.

In a previous post vectorization was recommended, but now I can not figure out how to vectorize when I have different indices operations.

## closed as off-topic by Toby Speight, Graipher, IEatBagels, Sᴀᴍ Onᴇᴌᴀ, LudisposedFeb 28 at 14:39

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I was the one who recommended vectorizing. I think the first approach would be to convert your code to using 2D indexes. I think this will make the vectorizing clearer:

for j in range(rows):
for i in range(cols):
i1 = i+1 if i<cols-1 else i-1
i2 = i-1 if i>0 else i+1
j1 = j+1 if j<rows-1 else j-1
j2 = j-1 if j>0 else j+1

w1 = U[j, i]
w2 = U[j, i-1] if i > 0 else U[j, i]
w3 = V[j, i]
w4 = V[j-1, i] if j > 0 else V[j, i]

p = parray[j, i]
p1 = parray[j, i1]
p2 = parray[j, i2]
p3 = parray[j1, i]
p4 = parray[j2, i]

zarray[j, i] = (w1 + w2 + w3 + w4)*p - (w1*p1 + w2*p2 + w3*p3 + w4*p4)


So, if I am reading this code right, you are shifting some rows and columns around. So the next step is to re-implement this by making copies of the arrays that follow the same patterns:

U1 = np.empty_like(U)
V1 = np.empty_like(V)
parray1 = np.empty_like(parray)
parray2 = np.empty_like(parray)
parray3 = np.empty_like(parray)
parray4 = np.empty_like(parray)

U1[:, 1:] = U[:, :-1]  # this is w2
V1[1:, :] = V[:-1, :]  # this is w4
U1[:, 0] = U[:, 0]
V1[0, :] = V[0, :]

parray1[:, :-1] = parray[:, 1:]  # this is the result of k1
parray2[:, 1:] = parray[:, :-1]  # this is the result of k2
parray3[:-1, :] = parray[1:, :]  # this is the result of k3
parray4[1:, :] = parray[:-1, :]  # this is the result of k4

parray1[:, -1] = parray[:, -2]
parray2[:, 0] = parray[:, 1]
parray3[-1, :] = parray[-2, :]
parray4[0, :] = parray[1, :]

zarray = (U+U1+V+V1)*parray - (U*parray1 + U1*parray2 + V*parray3 + V1*parray4)


You could also use np.hstack and np.vstack instead of slices:

U1 = np.hstack([U[:, :1], U[:, :-1]])
V1 = np.vstack([V[:1, :], V[:-1, :]])

parray1 = np.hstack([parray[:, 1:], parray[:, -2:-1]])
parray2 = np.hstack([parray[:, 1:2], parray[:, :-1]])
parray3 = np.vstack([parray[1:, :], parray[-2:-1, :]])
parray4 = np.vstack([parray[1:2, :], parray[:-1, :]])

zarray = (U+U1+V+V1)*parray - (U*parray1 + U1*parray2 + V*parray3 + V1*parray4)

• Yes, this is the way I just solved! Thank you a lot! I thought (w1+w2+w3+w4) would be a inconvenience, but it just was the same as de previuos one. – FacundoGFlores Jun 23 '15 at 19:46
• Just an observation: which is a better way for copying? a[:] = v or a = v.copy()? – FacundoGFlores Jun 23 '15 at 19:48
• a[:] = v only works if a is already defined, so a = v.copy() is better for copying the whole thing. I guess you could pre-define the arrays using np.empty_like rather than by making copies, then slice those. I am not sure whether that would save much time or not. – TheBlackCat Jun 23 '15 at 19:56
• It turns out using np.empty_like is orders of magnitude faster, so I updated my answer to use it instead. – TheBlackCat Jun 23 '15 at 20:01
• I have also added an approach using hstack and vstack. I can't tell which is faster, my benchmarks are all over the place. – TheBlackCat Jun 23 '15 at 20:09